Composite cylinder by shrink fitting

Data

Inner tube: $a = 30$ mm, $r = 50$ mm. Outer tube: $r = 50$ mm, $b = 70$ mm. Radial interference $\delta = 0.06/2 = 0.03$ mm. Both tubes are steel, $E = 200\times10^3$ MPa.

        outer tube (50 to 70 mm)
     .--------------------.
    /   .--------------.   \
   |   /  inner tube    \   |
   |  |   (30 to 50)     |  |
   |   \                /   |
    \   '--------------'   /
     '--------------------'
        contact pressure p at r = 50

Contact pressure

After assembly, a pressure $p$ acts on the contact surface. The inner tube has external pressure $p$ only, the outer tube has internal pressure $p$ only. The radial interference equals the sum of the outward movement of the outer tube's inner surface and the inward movement of the inner tube's outer surface.

For a tube with internal pressure $p$ (inner radius $r$, outer radius $b$): hoop stress at inner surface $= p\dfrac{b^2 + r^2}{b^2 - r^2}$, so

$$u_{outer} = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \nu\right]$$

For a tube with external pressure $p$ only (inner radius $a$, outer radius $r$): hoop stress at outer surface $= -p\dfrac{r^2 + a^2}{r^2 - a^2}$, so the inward movement is

$$u_{inner} = \frac{p\,r}{E}\left[\frac{r^2 + a^2}{r^2 - a^2} - \nu\right]$$

Sum (the Poisson terms cancel for the same material):

$$\delta = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \frac{r^2 + a^2}{r^2 - a^2}\right]$$ $$\frac{b^2 + r^2}{b^2 - r^2} = \frac{4900 + 2500}{2400} = 3.0833,\qquad \frac{r^2 + a^2}{r^2 - a^2} = \frac{2500 + 900}{1600} = 2.1250$$ $$p = \frac{\delta E}{r\,(3.0833 + 2.1250)} = \frac{0.03\times200\times10^3}{50\times5.2083} = 23.04\ \text{MPa}$$

(b) Hoop stresses

Outer tube (internal pressure $p$):

$$\sigma_\theta(r=50) = p\,\frac{b^2 + r^2}{b^2 - r^2} = 23.04\times3.0833 = 71.04\ \text{MPa (tension)}$$ $$\sigma_\theta(r=70) = \frac{2pr^2}{b^2 - r^2} = \frac{2\times23.04\times2500}{2400} = 48.00\ \text{MPa (tension)}$$

Inner tube (external pressure $p$):

$$\sigma_\theta(r=50) = -p\,\frac{r^2 + a^2}{r^2 - a^2} = -23.04\times2.125 = -48.96\ \text{MPa}$$ $$\sigma_\theta(r=30) = -\frac{2pr^2}{r^2 - a^2} = -\frac{2\times23.04\times2500}{1600} = -72.00\ \text{MPa (compression)}$$

The radial stress is $-p = -23.04$ MPa at the contact surface and zero on the free surfaces.

(c) Heating of the outer tube

To slip the outer tube over, its bore must expand by the diametral interference $0.06$ mm (a small extra clearance is used in practice). The bore of 100 mm diameter expands by $\alpha\,d\,\Delta T$:

$$\Delta T = \frac{0.06}{12\times10^{-6}\times100} = 50\ ^\circ\text{C}$$

Answer: Contact pressure $= 23.04$ MPa; hoop stresses: outer tube $71.0$ MPa (bore) and $48.0$ MPa (outside); inner tube $-49.0$ MPa (outside) and $-72.0$ MPa (bore); outer tube to be heated by at least $50^\circ$C above the inner tube.